The zeroes of the quadratic polynomial x^{2}+ | vedclass

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Question

(A) Let the given quadratic polynomial be $p(x) = x^{2} + 99x + 127$.Comparing $p(x)$ with the standard form $ax^{2} + bx + c$,we get $a = 1$,$b = 99$

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both negative

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(A) Let the given quadratic polynomial be $p(x) = x^{2} + 99x + 127$.Comparing $p(x)$ with the standard form $ax^{2} + bx + c$,we get $a = 1$,$b = 99$,and $c = 127$.For a quadratic polynomial $ax^{2} + bx + c$,if $a, b, c > 0$,then the product of zeroes $\alpha \beta = \frac{c}{a} = 127 > 0$ and the sum of zeroes $\alpha + \beta = -\frac{b}{a} = -99 Since the product of the zeroes is positive,both zeroes must have the same sign.Since the sum of the zeroes is negative,both zeroes must be negative.Alternatively,using the quadratic formula $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$:$x = \frac{-99 \pm \sqrt{99^{2} - 4(1)(127)}}{2(1)} = \frac{-99 \pm \sqrt{9801 - 508}}{2} = \frac{-99 \pm \sqrt{9293}}{2}$.Since $\sqrt{9293} \approx 96.4$,the roots are approximately $\frac{-99 + 96.4}{2} = -1.3$ and $\frac{-99 - 96.4}{2} = -97.7$.Thus,both zeroes are negative.

The zeroes of the quadratic polynomial $x^{2}+99x+127$ are

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